Simplify and expand the following expression: $ \dfrac{1}{2n + 2}+ \dfrac{4}{3n - 27}+ \dfrac{3}{n^2 - 8n - 9} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2n + 2} = \dfrac{1}{2(n + 1)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{4}{3n - 27} = \dfrac{4}{3(n - 9)}$ We can factor the quadratic in the third term: $ \dfrac{3}{n^2 - 8n - 9} = \dfrac{3}{(n + 1)(n - 9)}$ Now we have: $ \dfrac{1}{2(n + 1)}+ \dfrac{4}{3(n - 9)}+ \dfrac{3}{(n + 1)(n - 9)} $ The least common multiple of the denominators is: $ 6(n + 1)(n - 9)$ In order to get the first term over $6(n + 1)(n - 9)$ , multiply by $\dfrac{3(n - 9)}{3(n - 9)}$ $ \dfrac{1}{2(n + 1)} \times \dfrac{3(n - 9)}{3(n - 9)} = \dfrac{3(n - 9)}{6(n + 1)(n - 9)} $ In order to get the second term over $6(n + 1)(n - 9)$ , multiply by $\dfrac{2(n + 1)}{2(n + 1)}$ $ \dfrac{4}{3(n - 9)} \times \dfrac{2(n + 1)}{2(n + 1)} = \dfrac{8(n + 1)}{6(n + 1)(n - 9)} $ In order to get the third term over $6(n + 1)(n - 9)$ , multiply by $\dfrac{6}{6}$ $ \dfrac{3}{(n + 1)(n - 9)} \times \dfrac{6}{6} = \dfrac{18}{6(n + 1)(n - 9)} $ Now we have: $ \dfrac{3(n - 9)}{6(n + 1)(n - 9)} + \dfrac{8(n + 1)}{6(n + 1)(n - 9)} + \dfrac{18}{6(n + 1)(n - 9)} $ $ = \dfrac{ 3(n - 9) + 8(n + 1) + 18} {6(n + 1)(n - 9)} $ Expand: $ = \dfrac{3n - 27 + 8n + 8 + 18}{6n^2 - 48n - 54} $ $ = \dfrac{11n - 1}{6n^2 - 48n - 54}$